﻿ A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. from Physics System of Particles and Rotational Motion Class 11 Manipur Board

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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

The free body diagram of the bar is shown in the following figure.

Length of the bar, l = 2 m

Tand T2 are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have

T1 Sin 36.90 = T2 Sin 53.1

=

⇒      T1 =  T2
For rotational equilibrium, on taking the torque about the centre of gravity, we have

T1 (Cos 36.9) × d = T2 Cos 53.1 (2 - d)

T1 × 0.800 d = T2 × 0.600 (2 - d

× T2 × 0.800d = T2 (0.600 × 2 - 0.600d

1.067d + 0.6d = 1.2

∴                         d = 1.2 / 1.67

= 0.72 m

Hence, the centre of gravity of the given bar lies 0.72 m from its left end.
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Is it necessary for centre of mass to lie within the body?

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The position of the centre of mass is calculated using the usual Newtonian type of equations of motion.
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What is the need of centre of mass?

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The concept of centre of mass of a system enables us to discuss overall motion of the system by replacing the system by an equivalent single point object.
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Is it necessary that there should be matter at the centre of mass of system?

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What is the significance of defining the center of mass of a system?

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